When 1 mol `CrCl_(3).6H_(2)O` is treated with excess of `AgNO_(3)`, 3 mol of `AgCl` are obtained. The formula of the coplex is

Options

A. `[CrCl_(3)(H_(2)O)_(3)]. 3H_(2)O`
B. `[CrCl_(2)(H_(2)O)_(4)]Cl.2 H_(2)O`
C. `[CrCl (H_(2)O)_(5)]Cl_(2).H_(2)O`
D. `[Cr(H_(2)O)_(6)]Cl_(3)`

Answer

D

Explanation

1 mole of `AgNO_(3)` precipitates one free chlorids ion `(Cl^(-))`.
Here, 3 moles of `AgCl` are precipitated by excess of `AgNO_(3)`. Hence, there must be three free `Cl^(-)` ions.
So, the formula of the complex can be `[Cr(H_(2)O)_(6)]Cl_(3)` and correct choice is (d).